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Question 614
a.

Use the substitution u=2+sin⁡θu = 2 + \sin \thetau=2+sinθ to show that the integral

∫24sin⁡2θ(2+sin⁡θ)3dθ \int \frac{24 \sin 2\theta}{(2 + \sin \theta)^3} d\theta ∫(2+sinθ)324sin2θ​dθ

can be written in the form

∫(48u2−96u3)du \int \left( \frac{48}{u^2} - \frac{96}{u^3} \right) du ∫(u248​−u396​)du
[4]
b.

The torque τ\tauτ (in N m) generated by a mechanical component is modeled by the function

τ(θ)=24sin⁡2θ(2+sin⁡θ)3,0≤θ≤π2 \tau(\theta) = \frac{24 \sin 2\theta}{(2 + \sin \theta)^3}, \quad 0 \le \theta \le \frac{\pi}{2} τ(θ)=(2+sinθ)324sin2θ​,0≤θ≤2π​

where θ\thetaθ is the angle of rotation in radians. Calculate the exact value of the total work done, given by ∫0π/2τ(θ)dθ\int_{0}^{\pi/2} \tau(\theta) d\theta∫0π/2​τ(θ)dθ.

Show each stage of your working and give your answer as a fraction in its simplest form.

[4]
Markscheme

Integration Questions

  1. A Level
  2. /Maths
  3. /Integration

864 exam-style questions on Edexcel A Level Maths Integration, covering 11.1 Integrating Standard Functions, 11.2 Integrating f(ax + b), 11.3 Using Trigonometric Identities, 11.4 Reverse Chain Rule, 11.5 Integration by Substitution, 11.6 Integration by Parts, 11.7 Partial Fractions, 11.8 Finding Areas, 11.9 The Trapezium Rule, 11.10 Solving Differential Equations, and 11.11 Modelling with Differential Equations. Each one has a worked solution and a mark scheme showing where the marks go.

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